題意：

題意： 有n個(gè)人，m對(duì)人相互認(rèn)識(shí)； 問(wèn)能否分成兩個(gè)組，組內(nèi)任意兩個(gè)人之間不認(rèn)識(shí)； 若不能，則輸出No； 若能，則相互認(rèn)識(shí)的兩個(gè)人一間房，求最多需要幾間房；
給出一些學(xué)生的認(rèn)識(shí)情況，比如A和B認(rèn)識(shí)，B和C認(rèn)識(shí)，但是A和C不一定認(rèn)識(shí)。現(xiàn)在問(wèn)能否將這些學(xué)生分成兩個(gè)組，并且每組中的學(xué)生互相不認(rèn)識(shí)，如果能分，求出最大能匹配的學(xué)生對(duì)數(shù)。

題目

There are a group of students. Some of them may know each other, while others don’t. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don’t know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

Calculate the maximum number of pairs that can be arranged into these double rooms.

Input

For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

Proceed to the end of file.

Output

If these students cannot be divided into two groups, print “No”. Otherwise, print the maximum number of pairs that can be arranged in those rooms.

Sample Input

4 4
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6

Sample Output

No
3

題解：

1首先bfs判斷是否是二分圖，然后求二分最大匹配。
2判斷是否為二分圖：在無(wú)向圖G中，無(wú)向圖G為二分圖的充分必要條件是：G至少有兩個(gè)頂點(diǎn),且當(dāng)存在回路時(shí)，其所有回路的長(zhǎng)度均為偶數(shù)。回路就是環(huán)路，也就是判斷是否存在奇數(shù)環(huán)。如果存在奇數(shù)回路(回路中節(jié)點(diǎn)個(gè)數(shù)為奇數(shù))，則不是二分圖。否則是二分圖。
采用染色法+bfs（染色法是將一個(gè)點(diǎn)先染色，然后把和它相鄰的點(diǎn)染成不同的顏色，如果遇到相鄰點(diǎn)的顏色相同的情況就不是二分圖）
3 染色法判斷回路奇偶性：把相鄰兩點(diǎn)染成黑白兩色，如果相鄰兩點(diǎn)出現(xiàn)顏色相同則存在奇數(shù)回路。也就是非二分圖。
4匹配的對(duì)數(shù)：由于是兩個(gè)相同的集合進(jìn)行配對(duì)，所以最后將結(jié)果除2

AC代碼

#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; const int M=2e2+10; int n,m,dp[M],book[M],e[M],map[M][M]; bool dfs()//先要判斷能否分成兩組，使得每組內(nèi)的人互不認(rèn)識(shí)，即判斷是否為二分圖 {memset(dp,-1,sizeof(dp));//染色數(shù)組，-1為未染，0，1則為兩種不同顏色for(int i=1; i<=n; i++){if(dp[i]!=-1)continue;dp[i]=0;queue<int>q;q.push(i);while(!q.empty()){int head=q.front();q.pop();for(int j=1; j<=n; j++){if(!map[head][j])continue;if(dp[j]!=-1&&dp[head]==dp[j])//把相鄰兩點(diǎn)染成黑白兩色，如果相鄰兩點(diǎn)出現(xiàn)顏色相同則存在奇數(shù)回路。也就是非二分圖。return false;else if(dp[j]==-1){dp[j]=!dp[head];q.push(j);}}}}return true; } bool math(int x)//匈牙利算法 {for(int i=1; i<=n; i++)if(!book[i]&&map[x][i]){book[i]=1;if(!e[i]||math(e[i])){e[i]=x;return true;}}return false; } int main() {while(~scanf("%d%d",&n,&m)){memset(map,0,sizeof(map));for(int i=1; i<=m; i++){int a,b;scanf("%d%d",&a,&b);map[a][b]=map[b][a]=1;}if(dfs()==false){printf("No\n");continue;}int ans=0;memset(e,0,sizeof(e));for(int i=1; i<=n; i++){memset(book,0,sizeof(book));if(math(i))ans++;}printf("%d\n",ans/2);//求對(duì)數(shù)，除2}return 0; }

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